6
$\begingroup$

I believe there is a misconcept in my mind about the $\chi^2$ and/or standard normal distribution. Hence, I would like you to help me to understand what does it means that the $\chi_k^2$ distribution is a sum of $k$ independent, squared, standard, normal distributions.

In fact, according with many online sources, such as Wikipedia, the $\chi^2$ is

$$ \chi_k^2 = \sum_{i=1}^k Z_i^2 $$

where $Z_i$ are the $k$ different standard normal distributions. Furthermore, according with such sources, in some ways this sum becomes

$$ f(x;k) = \frac{ x^{ \frac{k}{2} - 1 }e^{ -\frac{x}{2} } }{ 2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right) }\quad\textrm{ if }x>0,\qquad 0\qquad\textrm{ otherwise} $$

My problem is in what follows:

As far as I know, the normal distribution is

$$ Z = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

and it gets "normalized" with a transformation that makes $\mu = 0$ and $\sigma = 1$, which is

$$ Z(0,1) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} $$

Now, the STANDARD normal distribution has no parameters at all, which means that in the $\chi_k^2$ distribution the sum could be replaced just with a multiplication

$$ \chi_k^2 = kZ^2 $$

which has to be wrong, otherwise there would be not such a complicated defition of the $\chi^2$ distribution.

What's wrong in all of this? Please, let me know

$\endgroup$
2
  • 4
    $\begingroup$ There is a difference in distribution between adding $n$ i.i.d. random variables and multiplying one of them by $n$. For example, the former has variance $n\sigma^2$ and the latter $n^2\sigma^2$. The shape can change too (unless they are "stable" - e.g. normal) $\endgroup$
    – Henry
    Commented Aug 19, 2022 at 8:06
  • $\begingroup$ in some ways this sum becomes... it happens in a legit way. You can take resort to the uniqueness of MGF to derive the distribution of a gamma variate with the concerned parameters. $\endgroup$ Commented Aug 19, 2022 at 8:14

1 Answer 1

14
$\begingroup$

Sampling one value from $$ \sum_{i=1}^k Z_i^2 $$ requires to make one draw from $Z_1$, one draw from $Z_2$, and so forth. In other words, you must make $k$ independent draws from the $N(0, 1)$ distribution.

On the other hand, sampling one value from $$ kZ^2 $$ requires to make one single draw from $Z$, square it, and to multiply it by $k$.


sample1 <- rnorm(n = 1e4)^2 + rnorm(n = 1e4)^2 + rnorm(n = 1e4)^2
sample2 <- 3 * rnorm(n = 1e4)^2


curve(dchisq(x, df = 3), from = 0, to = 40, col = "red", lwd = 2)
lines(density(sample1), col = "blue")
lines(density(sample2), col = "green")

enter image description here


from seaborn import displot
import numpy.random as dists
import pandas as pd

sample_size = 10**4
sample1 = dists.normal(size = sample_size)**2 + dists.normal(size = sample_size)**2 + dists.normal(size = sample_size)**2
sample2 = 3 * dists.normal(size = sample_size)**2
sample3 = dists.chisquare(df = 3, size = sample_size)

plot_data = pd.concat([pd.DataFrame({'label': '3 independent chi_sq_1',
                                     'data': sample1}),
                       pd.DataFrame({'label': '3 times chi_sq_1',
                                     'data': sample2}),
                       pd.DataFrame({'label': 'chi_sq_3',
                                     'data': sample3})],
                      ignore_index = True)

displot(data = plot_data, x = 'data', hue = 'label')
displot(data = plot_data, x = 'data', hue = 'label', kind = 'ecdf')

enter image description here enter image description here

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.